# Harm osc, anharmonic perturbation

1. Jan 2, 2008

### WarnK

1. The problem statement, all variables and given/known data
Particle bound by
$$V = \frac{1}{2} m \omega^2 x^2 - a x^3$$
for small x. Show that the mean position of the particle changes with the energy of the eigenstates when $$a$$ is small, so first order perturbation theory works.

2. Relevant equations
For the harmonic oscillator
$$x = \sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a)$$

3. The attempt at a solution
That x^3 perturbation will give an odd number of creation/destruction operators, so there's no shift in energy eigenvalues to first order in perturbation theory. But how does that help answering the question?

2. Jan 2, 2008

### Dr Transport

There is a first order transition, but it is between different energy states.

3. Jan 4, 2008

### WarnK

But how does that help answering the question?

4. Jan 12, 2008

### WarnK

Noone have any ideas on this?

5. Jan 22, 2008

### WarnK

I'm not going to give up on this! The question asks to calculate the mean position of the particle, so if I do that:
$$<x> = \sqrt{\frac{\hbar}{2m\omega}}<n|a^{\dagger}+a|n> = 0$$
That's just zero, haveing nothing to do with energy eigernstates or size of a?

Sure I could calculate the perturbated energy to second order, getting terms like
$$a^2 \big( \frac{\hbar}{2m\omega} \big)^3 <n|a^{\dagger} aa \frac{1}{E_n-H_0} a^{\dagger} a a^{\dagger} |n> = -n(n+1)^2 \frac{\hbar^2 a^2}{8m^3 \omega^4}$$
and so on, but I don't see how that has anyhting to do with the question asked.

6. Jan 22, 2008

### J.D.

Shouldn't you consider the expanded eigenkets?

$$| n > = |n^{(0)}>+\lambda |n^{(1)}> + \lambda^2 |n^{(2)}> + \dots$$

Then you get to first order something like:

$$<x> = < n^{(0)} | x | n^{(0)} > + \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2) = \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2)$$

Could that be the correct approach?

Last edited: Jan 22, 2008
7. Jan 22, 2008

### Aaronse_r

I think that you dont really have to solve for the corrected energy's. You need the new eigenkets, and then just solve for the expectation value, <x>, by taking the inner product of x and the eigenkets.